NCERT Solutions Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

NCERT Solutions for Class 12 Mathematics Chapter 2 is available on the Extramarks website for all kids who are preparing for class 12 board exams. This chapter is about Inverse Trigonometric Functions, and our solutions provide theoretical knowledge and answer solutions to all questions from the NCERT textbook. Our solution is built based on the CBSE NCERT latest 2022-2023 syllabus, and we offer a step-by-step guide for students to comprehend the concepts thoroughly. Students can take full advantage of Extramarks NCERT Solutions for Class 12 Mathematics Chapter 2 for their CBSE Class 12 Term one exam preparation.

The Inverse Trigonometric Functions chapter is an amalgamation of inverse function and trigonometry. It plays a vital role in Calculus concepts to define many integrals. One will learn about the restrictions on domains and ranges of trigonometric functions. Extramarks solutions cover theories, vital concepts, and a lot of solved examples for students to understand the topic easily. If you are looking for perfect study material, one must refer to Extramarks NCERT Solutions Class 12 Mathematics Chapter 2.

You can regularly visit the Extramarks website to check the latest updates and news about the CBSE exams. You can also download NCERT Solutions Class 9, NCERT Solutions Class 10, and NCERT Solutions 11, as those would be important for you to build a good foundation for Class 12 preparation.

Key Topics Covered In NCERT Solution for Class 12 Mathematics Chapter 2

To easily understand Class 12 Mathematics Chapter 2,  students need to know Chapter 1 on Relations and Functions properly. The underlying principles behind the theories and properties from earlier chapters will continue in Chapter 2. So students should understand functions and how to invert a function and solve questions based on it. The concepts of Inverse trigonometric functions covered in Chapter 2 forms the basis for calculus-based further topics. These concepts of inverse trigonometric functions would go a long way in many science and engineering-oriented curriculum. 

In Extramarks NCERT Solutions Class 12 Mathematics Chapter 2, students can expect all topics to be covered and explained in detail. It starts with a chapter introduction and includes sections on basic concepts, trigonometric functions, principle value and inverse function. Studying this chapter will strengthen the foundation of calculus, an important section for many competitive examinations such as NEET and JEE (Main and Advanced) exams. 

The key topics covered in chapter 2 include:

Exercise  Topic
2.1 Introduction
2.2 Basic Concepts
2.3 Properties of Inverse Trigonometric Functions
Others Miscellaneous Q&A

NCERT Solutions Class 12 Mathematics Chapter 2 requires students to use their critical thinking ability and apply a wide range of formulas they have learnt. 

2.1 Introduction

This chapter on  Inverse trigonometric functions is crucial in higher-grade Mathematics for many engineering and science streams as it builds the fundamentals of calculus. Inverses of trigonometric functions exist to overcome the restrictions we have on domains and their respective ranges. To solve inverse trigonometric functions, one must first understand the trigonometric ratios. The behaviour of the trigonometric functions is usually demonstrated via graphical methods.

2.2 Basic Concepts

This section will quickly revise the corresponding invertible functions. In addition, you will obtain a precise understanding of the functions of graphs and integers. Students will understand the concept that the graph of an inverse function can be extracted from the correlated graph of the original function as a mirror image. 

It will develop a better understanding between the domain and the co-domain. The domains are correlated to the principles of functions. This segment will help the students find the unknown angle's value. Inverse sine and inverse cosine are essential to study. And their basics and root concepts are mentioned in this segment. 

2.3 Properties of Trigonometry

This section will cover various essential properties of inverse trigonometric functions. These are the inverse functions of sine, cosine, tangent, cotangent, secant, and cosecant with restricted domains. It would teach students the properties and relationships that exist between various trigonometric functions via formulas.

Few of the formulas covered in the NCERT Solutions for Class 12 Mathematics chapter 2 are given below:

  •  sin-1 (–x) = – sin-1 x
  • tan-1 x + cot-1 x = π/2
  • sin-1 x + cos-1 x = π/2
  • cos-1 (–x) = π – cos-1 x
  • cot-1 (–x) = π – cot-1 x

Students should also refer to additional study resources such as Extramarks short chapter notes and solved solutions for NCERT Exemplars.

Download NCERT Solutions for Class 12 Mathematics Chapter 2 Exercise & Answer Solutions

NCERT Solutions for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions is available in a pdf format on the Extramarks website for free. It covers theoretical explanations and has step-by-step solutions to all questions from the NCERT textbook. As the CBSE Class 12 exams approaches, students should leverage the solution guide to get thorough with the Inverse Trigonometric Functions chapter.

In our solutions guide for the Class 12 Mathematics chapter, we have covered topics on domains and ranges of trigonometry-related functions that validate the inverse functions' existence and help observe their behaviour via graphical representations. Experts advise students to revise this chapter multiple times as it's a core topic in Calculus that helps define many integrals. And Calculus goes a long way in the advanced level Mathematics syllabus for engineering, architecture and science courses.

Click on the below links to download exercise specific questions and solutions for NCERT Solutions for Class 12 Mathematics Chapter 2:

  • Chapter 2: Exercise 2.1 Solutions: 14 Questions (12 Short Answers, 2 MCQ)
  • Chapter 2: Exercise 2.2 Solutions: 21 Questions (18 Short Answers, 3 MCQ)
  • Chapter 2: Miscellaneous Exercise Solutions: 17 Questions (14 Short Answers, 3 MCQ)

Along with Class 12 Mathematics solutions, you can explore NCERT solutions on our Extramarks website for all primary and secondary classes.

NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions for Class 9, Class 10, aand Class 11.

NCERT Exemplar Class 12 Mathematics PDF

NCERT Exemplar Class 12 Mathematics PDF is an excellent source of information for CBSE students preparing for their 12th standard exams. It consists of various examples with a detailed solution set. While revising from NCERT Exemplars, students gather complete knowledge on the Mathematics concepts and gain deeper insights on a variety of interlinked topics between various chapters. 

Exemplar books have proved beneficial for students in CBSE and other curriculum as well. It covers complex theories that confidently prepare students to face the upcoming examinations. Students appearing for Class 12 are advised to download the NCERT Exemplar Class 12 Mathematics pdf file and involve it in their core study materials. Students will get a deeper knowledge of solving complex queries, and it will help prepare them for boards and other entrance examinations. 

After referring to the NCERT Solutions and NCERT Exemplar, the students think logically about a problem. As a result, students can jump to more advanced and higher-level conceptual questions. By studying from the Exemplar, you can prepare well for upcoming entrance exams like JEE Main, JEE Advanced, NEET, and WBJEE. 

Key Features of NCERT Solutions Class 12 Mathematics Chapter 2

To attain good marks in the exam, students must revise all previous concepts. Therefore, NCERT Solutions for Class 12 Mathematics Chapter 2 offers a complete solution for all problems. The key features are provided: 

  • Extramarks NCERT Solutions Class 12 Mathematics Chapter 2 is prepared by Mathematics experts.
  • The solutions help to clear the student's doubts and obstacles by building the base and explaining the chapter's core concepts. 
  • With the Inverse Trigonometric Functions chapter solutions, students will be able to apply the concepts  required to solve many Calculus related problems.

Q.1 Find the principal values of the following



Let  sin112=x. Then, sinx=12.We know that the range of the principal value branch of sin1 isπ2,π2 and sin π6=12.Therefore, principal value of sin1 12=π6




Letcos132=x. Then, cosx=32.We know that the range of the principal value branch of cos1 is0, π and cosπ6=32.




Letcosec1(2)=x. Then, cosecx=2.We know that the range of the principal value branch of cosec1 isπ2,π2{0} and cosecπ6=2.Therefore, principal value of cosec1(2)=π6




Let  tan1(3)=x. Then, tanx=3.We know that the range of the principal value branch of tan1 isπ2,π2 and tan π3=3.Therefore, principal value of tan1(3)=π3.




Let  cos112=x. Then, cosx=12.We know that the range of the principal value branch of cos1 is[0,π] and cos π3=12.cosx=12          =cos π3          =cos ππ3          =cos 2π3x=2π3Therefore, principal value of cos112=2π3.




Let  tan1(1)=1.We  know  that  the  range  of  the  principle  value  branch  of  tan1  is  π2,π2  and  tanπ4=1.Therefore,principle  value  of  tan1(1)=π4




Letsec123=x. Then, secx=23.We know that the range of the principal value branch of cos1  is[0,π]π2 and secπ6=23.Therefore, principal value of sec123=π6




Let  cot13=x. Then, cotx=3.We know that the range of the principal value branch of cot1 is0,π and cotπ6=3.Therefore, principal value of cot13=π6.




Let  cos1 12=x. Then, cosx=12.We know that the range of the principal value branch of cos1 is[0,π] and cosπ4=12.cosx=12          =cosπ4          =cosππ4          =cos 3π4x=3π4Therefore, principal value of cos112=3π4.




Letcosec1(2)=x. Then, cosecx=2.We know that the range of the principal value branch of cosec1 is π2,π2{0} and cosecπ4=2.Therefore, principal value of cosec1(2)=π4

Q.11 Find the values of the following:

tan1(1)+cos1 12+sin1 12


Given: tan1(1)+ cos1 12+ sin1 12We shall find the principal value of each term as given below:Let  tan1(1)=x. Then, tanx=1.We know that the range of the principal value branch of tan1 isπ2,π2 and t an π4=1.Therefore, principal value of tan1(1)=π4.second term:Let  cos1 12=x. Then, cosx=12.We know that the range of the principal value branch of cos1 is[0,π] and cos π3=12.cosx=12          =cos π3          =cos ππ3          =cos 2π3and third term:Let  sin1 12=x. Then, sinx=12.We know that the range of the principal value branch of sin1 isπ2,π2 and sin π6=12.Therefore, principal value of sin1 12=π6Then,tan1(1)+ cos112+ sin112=π4+2π3π6  =3π+8π2π12  =9π12  =3π4.


cos112+2sin1 12


cos112 +2sin1 12First term: cos1 12Letcos1 12=x. Then, cosx=12.We know that the range of the principal value branch of cos1 is[0,π] and cos π3=12.Therefore, principal value of cos1 12=π3Second term:  2sin1 12Let  sin112=x. Then, sinx=12.We know that the range of the principal value branch of sin1 isπ2,π2 and sin π6=12.Therefore, principal value of 2sin112=2×π6=π3cos1 12+2sin112=π3+π3      =2π3


Ifsin1x=y, thenA)0yπB)π2yπ2C) 0<y<πD)π2<y<π2


Since, sin1: [1, 1]π2,π2So, option B is the correct answer.


tan13sec1(2) is equal to(A) π(B) π3(C) π3(D) 2π3


Given:    tan13sec1(2)Since,  tan13=π3, andlet  sec1(2)=x. Then, secx=2.We know that the range of the principal value branch of sec1 is[0,π]π2 and secπ3=2. secx=2=sec π3    =sec ππ3    =sec2π3Therefore, principal value of sec1(2)=2π3.    tan13sec1(2)=π32π3=π2π3=π3.Option  B is the correct answer.

Q.15 Prove the following:

3sin1x=sin1(3x4x3), x 12,12


R.H.S.=sin1(3x4x3)  =sin1(3sinθ4sin3θ) [Putting x= sinθ]  =sin1(sin3θ) [sin3x=3sinx4sin3x]  =3θ  =3sin1x [x=sinθθ=sin1x]  =R.H.S.


3cos1x=cos1 4x33x, x 12,1


R.H.S.=cos1(4x33x)=cos1(4cos3θ3cosθ) [Putting x= cosθ]=cos1(cos3θ) [cos3x=4cos3x3cosx]=3θ=3cos1x [x=cosθθ=cos1x]=R.H.S.


tan1211+tan1 724=tan1 12


L.H.S. =tan1211 + tan1 724    =tan1 211+7241211×724 tan1x+tan1y=tan1x+y1xy    =tan1 48+7711×2411×241411×24    =tan1125250    =tan1 12  =R.H.S.


2tan1 12+tan1 17=tan1 3117


L.H.S=2tan1 12+tan117  =tan12×121(12)2+ tan1 17 2tan1x=tan12x1x2  =tan143+tan1 17  =tan1 43+17143×17 tan1x+tan1y=tan1x+y1xy  =tan1 3117  =R.H.S.

Q.19 Write the following functions in the simplest form:

tan11+x21x, x0


Given:tan11+x21x=tan11+tan2θ1tanθ[Putting x=tanθ]    =tan1sec2θ1tanθ[1+tan2θ=sec2θ]    =tan1secθ1tanθ    =tan11cosθsinθ    =tan1 tanθ2[tan1(tanx)=x]    =θ2    =12tan1x


tan11x21, |x|>1


tan11x21, |x|>1


ta n 1 ( 1cosx 1+cosx ), x<π MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWF0bGaa8xyaiaa=5gadaahaaWcbeqaaiaa=1cacaWFXaaaaOWaaeWaaeaadaGcaaqaamaalaaabaGaa8xmaiaa=1cacaWFJbGaa83Baiaa=nhacaWF4baabaGaa8xmaiaa=TcacaWFJbGaa83Baiaa=nhacaWF4baaaaWcbeaaaOGaayjkaiaawMcaaiaa=XcacaWLjaGaaCzcaiaa=HhacqGH8aapcqaHapaCaaa@4EDC@


tan -1 ( 1-cosx 1+cosx )= tan -1 ( 2sin 2 ( x 2 ) 2cos 2 ( x 2 ) ) [ cos2x=12 sin 2 x =2 cos 2 x1 ] = tan 1 ( tan x 2 ) = x 2 [ tan 1 ( tanx )=x ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E4FD@


tan1cosxsinxcosx+sinx, 0<x<π


tan1cosx sinxcosx +sinx=tan11tanx1+tanx Dividing numerator and denominator by sinx    =tan1tanπ4x tan π4x=1tanx1+tanx     =π4x[tan1(tanx)=x]


tan1xa2x2, |x|<a


tan1xa2x2=tan1asinθa2a2sin2θ[Putting x=a sin θ]      =tan1 asinθacosθ [1sin2θ=cos2θ]      =tan1(tanθ) sinθcosθ=tanθ      =θ      =sin1xa [x=asinθ]


tan1 3a2xx3a33ax2, a>0; a3xa3


tan1 3a2xx3a33ax2=tan1 3a3tanθ(atanθ)3a33a(atanθ)2 [Putting x=atanθ]      =tan1 3a3tanθa3tan3θa33a3tan2θ      =tan1 a3(3tanθtan3θ)a3(13tan2θ)      =tan1(tan3θ) tan3θ=3tanθtan3θ13tan2θ      =3θ      =3tan1xa [x=atanθ]

Q.25 Find the values of each of the following:

tan1 2cos 2sin112


tan1 2cos 2sin112 tan1 2cos 2×π6        =tan12cosπ3        =tan12×12        =tan1(1)=π4


cot tan1a+cot1a


cot tan1a+cot1a=cot tan1x+cot1x=π2 π2 =0


tan12 sin12x1+x2+cos11y21+y2 , |x|<1, y>0 and xy <1


tan12 sin12x1+x2+cos11y21+y2=tan12 sin12tanα1+tan2α+cos11tan2β1+tan2β [Putting x=tanα, y=tanβ]=tan12 sin1(sin2α)+cos1(cos2β)=tan12(2α+2β)=tan(α+β)=tanα+tanβ1tanαtanβ=x+y1xy


If sin sin115+cos1x=1, then find the value of x.


sin sin115+cos1x=1sin sin115+ cos1x =sin1π2          sin115+cos1x=π2            sin115=π2cos1x            sin115=sin1x sin1x+cos1x=π2    15=x


If tan1x1x2 + tan1x+1x+2=π4, then find the value of x.


        tan1x1x2+tan1x+1x+2=π4 tan1x1x2+x+1x+21x1x2 x+1x+2=π4 tan1x+tan1y=tan1x+y1xy  (x2+x2+x2x2(x2)(x+2)x24x2+1(x2)(x+2))=tanπ4            2x243=1  2x24=3    x=±12


sin1 sin2π3


sin1 sin2π3 =sinRange of sinθ isπ2,π2 1 sinππ3 =sin1sinπ3sin(πθ)=sinθ=π3


tan1 tan3π4


tan1tan3π4=tan1tan ππ4 Range of tanθ isπ2,π2 =tan1 tanπ4 [tan(πθ)=tanθ]=tan1 tanπ4 [tan(θ)=tanθ]=π4


tan sin135+cot132


tan sin135+cot132Let sin135=x  and  cot132=y  sinx=35  and  coty=32cosx=1sin2x        =1352=45andtan y=1cot y  =132=23tanx=sinxcosx=3545  =34x=tan1 34  andy=tan1 23sin135=tan1 34  and  cot132=tan1 23tan sin135+cot132=tan tan1 34+tan123=tan tan134+23134×23=tan tan1176=176


cos1 cos7π6 is equal to(A) 7π6(B) 5π6(C) π3(D) π6


cos1 cos7π6=cos1 cos 2π5π6 [Range of cosθ is[0,π]]=cos1 cos 5π6 [cos (2πθ)=cosθ]=5π6 [cos1(cosx)=x]Hence option (b) is correct.


sin π3sin1 12 is equal to(A)12(B)13(C)14(D)1


sinπ3sin112=sin π3+sin112 [sin(θ)=sinθ]=sin π3+π6 sinπ6=12=sin 2π+π6=sin 3π6=sin π2=1 sinπ2=1Hence, option (D)  is correct.


tan13cot13 isequaltoAπBπ2C0D23


tan13cot1(3)=tan1(3)cot1(3)[tan(θ)=tanθ]={tan1(3)+cot1(3)}=π2[tan1x+con1x=π2]Hence option (B) is correct.


Find the value of cos1 cos13π6.


Range of cosθ  is[0, π]cos1 cos13π6=cos1 cos2π+π6      =cos1 cosπ6 [cos(2π+θ)=cos θ]      =π6


Find the value of tan1 tan7π6.


Range of tanθ  isπ2,π2tan1 tan7π6 = tan1 tan π+π6      =tan1tanπ6 [tan(π+θ)=tanθ]      =π6


Prove that:2sin1 35=tan1247


Let sin135=xsinx=35and  cosx =1sin2x=1 352=45tanx=sinxcosx =3545 =34Since,tan 2x =2 tan x1+tan2x =2×341 342 =2417   2x = tan1 24172 sin1 35=tan1 2417


Prove that:sin1817 + sin1 35=tan17736


sin1 817 + sin1 35=tan17736Let sin1 817=xsinx=817and  cosx=1sin2x=1(817)2=1517        tanx=sinxcosx=8171517=815          x=tan1815sin1 817=tan1815Again,let  sin1(35)=ysiny=35so,        tany=siny1sin2y    =351(35)2    =34y=tan1 34  sin1 35=tan1 34L.H.S=sin1 817+si n1 35=tan1 815 + tan1 34=tan1 815+341815×34=tan1 7736=R.H.S.


Prove that:cos1 45+ cos1 1213 =cos13365


Let cos1 45= x and cos1 1213=ycosx=45 and cosy=1213sinx=1cos2x  and    siny=1cos2ysinx=1452  and    siny=112132sinx=35  and    siny=513Then,cos(x+y)=cosx cosysinx siny  =45×121335×513  =48155×13x+y=cos13365cos1 45+ cos1 1213=cos1 3365.


Prove that:cos11213+ sin1 35=sin15665


Letcos1 1213=x and sin1 35=ycosx=1213  and  siny=35sinx=1cos2x  and  cosy=1sin2ysinx=112132  and  cosy=1352sinx=513  and  cosy=45Since,sin(x+y)=sinx cosy+siny cosx  =513 45 + 35 1213  =20+3665=5665    x+y=sin1 5665cos11213 +sin1 35=sin1 5665.


Prove that:tan16316=sin1 513+ cos135


Let sin1513=x and cos135=ysinx=513  and  cosy=35  cosx=1sin2x  and  siny=1cos2y  cosx=1(513)2  and  siny=1(35)2  cosx=1213  and      siny=45tanx=sinxcosx=512  and  tany=43Since      tan(x+y)=tanx+tany1tanxtany=512+431512×43      x+y=tan1(6316)sin1 513+cos135=tan1 6316


Prove that:tan1(15)+tan1(17)+tan1(13)+tan1(18)=π4


L.H.S.=tan1(15)+tan1(17)+tan1(13)+tan1(18)    =tan1(15+17115×17)+tan1(13+18113×18)[tan1x+ tan1 y=tan1(x+y1xy)]    =tan1(1234)+tan1(1123)    =tan1(1234+112311234×1123)    =tan1(650650)    =tan1(1)    =π4=R.H.S.


Prove that:tan1x=12cos1(1x1+x),  x [0,1]


R.H.S.=12cos11x1+x      =12cos1 1tan2θ1+tan2θ [Putting x = tan2θ]      =12cos1(cos2θ)      =2θ      = 2 tan1x [since,tanθ = xθ= tan1x]      =L.H.S.


Prove that:cot1(1+sinx+1sinx1+sinx1sinx)=x2,  x (0,π4)


L.H.S.=cot1(1+sinx+1sinx1+sinx1sinx)=cot1(sin2x2+cos2x2+2sinx2cosx2+sin2x2+cos2x22sinx2cosx2sin2x2+cos2x2+2sinx2cosx2sin2x2+cos2x22sinx2cosx2)=cot1((sinx2+cosx2)2+(cosx2sinx2)2(sinx2+cosx2)2(cosx2sinx2)2)=cot1(sinx2+cosx2+ cosx2sinx2sinx2+cosx2(cosx2sinx2))=cot1(2cosx22sinx2)=cot1(cotx2)=x2=R.H.S.


Prove that:tan1(1+x+1x1+x1x)=π412cos1x,    12x1[Hint:Putx=cos2θ]


L.H.S.=tan1(1+x+1x1+x1x)    =tan1(1+cos2θ+1cos2θ1+cos2θ1cos2θ)[Putting x=cos2θ]    =tan1(2cos2θ+2sin2θ2cos2θ2sin2θ)    =tan1(cosθ+sinθcosθsinθ)    =tan1(1+sinθcosθ1sinθcosθ)    =tan1(1+tanθ1tanθ)    =tan1tan(π4θ)=π4θ=R.H.S.


Prove that:9π894sin1(13)=94sin1(223)


L.H.S.=9π894sin1(13)  =94{π2sin1(13)}  =94cos1(13)...(i)Now, let x=cos1(13)cosx=13sinx=1cos2x        =1(13)2        =223          x=sin1(223)cos1(13)=sin1(223)So,from eq(i), we getL.H.S.=94cos1(13)=94sin1(223)=R.H.S.


Prove that:tan1(1+x+1x1+x1x)=π412cos1x,    12x1[Hint:Putx=cos2θ]


L.H.S.=tan1(1+x+1x1+x1x)    =tan1(1+cos2θ+1cos2θ1+cos2θ1cos2θ)[Putting x=cos2θ]    =tan1(2cos2θ+2sin2θ2cos2θ2sin2θ)    =tan1(cosθ+sinθcosθsinθ)    =tan1(1+sinθcosθ1sinθcosθ)    =tan1(1+tanθ1tanθ)    =tan1tan(π4θ)=π4θ=R.H.S.


Prove that:9π894sin1(13)=94sin1(223)


L.H.S.=9π894sin1(13)  =94{π2sin1(13)}  =94cos1(13)...(i)Now, let x=cos1(13)cosx=13sinx=1cos2x        =1(13)2        =223          x=sin1(223)cos1(13)=sin1(223)So,from eq(i), we getL.H.S.=94cos1(13)=94sin1(223)=R.H.S.

Q.50 Solve 2tan-1(cosx) = tan-1(2cosecx)


2tan-1(cosx)=tan-1(2cosecx)tan-1(2cosx1cos2x)=tan-1(2sinx)[2tan1x=tan1(2x1x2)]2cosx1cos2x=2sinx  cosxsin2x=1sinx  cosx=sinx    1=tanx    tanπ4=tanx   x=π4


Solve:tan11x1+x=12tan1x, (x>0)


                tan11x1+x=12tan1x      2tan11x1+x=tan1xtan1{2(1x1+x)1(1x1+x)2}=tan1x2(1x1+x)1(1x1+x)2=x    2(1x)(1+x)(1+x)2(1x)2=x          2(1x2)4x=x  1x2=2x2          1=3x2x=±13


sintan1x, x<1 is equal toax1x2b11x2c11+x2dx1+x2


sin( tan -1 x ) Let tan -1 x=θx=tanθ sec 2 θ=1+ tan 2 θ 1 cos 2 θ =1+ x 2 cos 2 θ= 1 1+ x 2 Since, sinθ= 1 cos 2 θ = 1 1 1+ x 2 = 1+ x 2 1 1+ x 2 = x 1+ x 2 θ= sin 1 ( x 1+ x 2 ) tan -1 x= sin 1 ( x 1+ x 2 ) Therefore, sin( tan -1 x )=sin{ sin 1 ( x 1+ x 2 ) } = x 1+ x 2 Thus,option ‘D’ is correct solution. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@499C@


sin1(1x)2sin1x=π2, then x is equal to(A) 0,12  (B   1,12(C) 0(D) 12


sin -1 ( 1x ) 2sin -1 x= π 2 1x=sin( π 2 +2 sin 1 x ) =cos( 2 sin 1 x ) [ sin( π 2 +x )=cosx ] =cos{ cos 1 ( 12 x 2 ) } ( Let sin 1 x=θ and cos2θ=1 2sin 2 x 2θ= cos 1 ( 12 sin 2 x ) 2 sin 1 x= cos 1 ( 1 x 2 ) ) 1x=12 x 2 2 x 2 x=0 x( 2x1 )=0 x=0or 2x1=0 x=0or x= 1 2 But on substitution of x= 1 2 L.H.S.= sin -1 ( 1-x ) 2sin -1 x = sin 1 ( 1 1 2 ) 2sin -1 ( 1 2 ) = π 6 2( π 6 ) = π 6 π 2 R.H.S. x= 1 2 is not the solution of given equation. Thus, x=0 is the solution of given equation. Therefore, option ‘C’ is correction solution of given equation. 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tan1xytan1xyx+y is equal toA π2B π3C π4D 3π4


tan1(xy) tan1(xyx+y)=tan1(xy) tan1(xy1xy+1)=tan1(xy) tan1{(1xy)(1+xy)}=tan1(xy) + tan1{(1xy)(1+xy)} [tan1(x)=tan1x]=tan1(xy) + tan1(1) tan1(xy)=tan1(1)=π4Therefore, option C is correct solution.

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